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Question 52436: Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,…to find the following:
a) What is r, the ratio between 2 consecutive terms?
Answer:
Show work in this space.
b) Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 10 terms? Please round your answer to 4 decimals.
Answer:
Show work in this space.
c) Using the formula for the sum of the first n terms of a geometric series, what is the sum of the first 12 terms? Please round your answer to 4 decimals.
Answer:
Show work in this space.
d) What observation can make about these sums? In particular, what number does it appear that the sum will always be smaller than?
Answer:
Found 2 solutions by AnlytcPhil, funmath:
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website! Use the geometric sequence of numbers 1, 1/2, 1/4, 1/8,…to find
the following:
a) What is r, the ratio between 2 consecutive terms?
Answer:
Show work in this space.
2nd term divided by 1st term = 1/2 ÷ 1 = 1/2
3rd term divided by 2nd term = 1/4 ÷ 1/2 = 1/4 × 2/1 = 2/4 = 1/2
4th term divided by 3rd term = 1/8 ÷ 1/4 = 1/8 × 4/1 = 4/8 = 1/2
Those all have to be equal the same for it to be a geometric
sequence. They are so it is. That common ratio is therefore the
number r = 1/2
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b) Using the formula for the sum of the first n terms of a
geometric series, what is the sum of the first 10 terms? Please
round your answer to 4 decimals.
Answer:
Show work in this space.
the sum of the first n terms is given by the formula
a1(1 - rn)
Sn = ------------
1 - r
where a1 = the first term or 1. r = 1/2, and n = 10
1[1 - (1/2)10]
S10 = -----------------
1 - 1/2
1 - (1/2)10
S10 = -----------------
1/2
1 - 1/210
S10 = -----------------
1/2
Multiply top and bottom by 2
2(1 - 1/210)
S10 = ----------------
2(1/2)
2 - 2/210
S10 = ----------------
1
S10 = 2 - 1/29
S10 = 2 - 1/512
S10 = 1024/2 = 1/512
S10 = 1023/512 = 1.998046875, or 1.9980 to
4 decimals
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c) Using the formula for the sum of the first
n terms of a geometric series, what is the sum
of the first 12 terms? Please round your answer
to 4 decimals.
Answer:
Show work in this space.
a1(1 - rn)
Sn = ------------
1 - r
where a1 = the first term or 1. r = 1/2, and
this time n = 12
1[1 - (1/2)12]
S12 = -----------------
1 - 1/2
1 - (1/2)12
S12 = -----------------
1/2
1 - 1/212
S12 = -----------------
1/2
Multiply top and bottom by 2
2(1 - 1/212)
S12 = ----------------
2(1/2)
2 - 2/212
S12 = ----------------
1
S12 = 2 - 1/211
S12 = 2 - 1/2048
S12 = 4096/2 = 1/2048
S12 = 4095/2048 = 1.9999511719, or 1.9999 to
4 decimals
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d) What observation can make about these sums?
In particular, what number does it appear that
the sum will always be smaller than?
Answer:
1.9980 and 1.9999 are very close to 2
If we just leave the n without replacing it in
the formula,
a1(1 - rn)
Sn = ------------
1 - r
where a1 = the first term or 1. r = 1/2, and
n = 10
1[1 - (1/2)n]
Sn = -----------------
1 - 1/2
1 - (1/2)n
Sn = -----------------
1/2
1 - 1/2n
Sn = -----------------
1/2
Multiply top and bottom by 2
2(1 - 1/2n)
Sn = -------------
2(1/2)
2 - 2/2n
Sn = ----------
1
Sn = 2 - 1/2n-1
The fraction 1/2n-1 is subtracted from 2 in the
formula. It gets smaller and smaller as n gets
larger because its denominator 2n-1 gets larger
and larger. So we are each time subtracting less
and less from 2. But the fraction will never
reach 0, so Sn will always be just a tiny
fraction under 2
Edwin
Answer by funmath(2933)  (Show Source):
You can put this solution on YOUR website! a) a(sub2)/a(sub1)=a(sub3)/a(sub2)=...=r
a(sub1)=1;a(sub2)=1/2
r=(1/2)/1=1/2
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b) S(subn)=a(sub1)((1-r^n)/(1-r))
n=10;a(sub1)=1;r=1/2
S(sub10)=(1)((1-(1/2)^10)/(1-1/2))
=(1-1/1024)/(1-1/2)
=(1023/1024)/(1/2)
=(1023/1024)*(2/1)
=1023/512
=1.9980
----------------------------------------
c)S(subn)=a(sub1)((1-r^n)/(1-r))
n=12;a(sub1)=1;r=1/2
S(sub12)=(1)(1-(1/2)^12)/(1-1/2)
=(1-1/4096)/(1-1/2)
=(4095/4096)/(1/2)
=(4095/4096)*(2/1)
=4095/2048
=1.9995
-----------------------------------------
d) It's getting closer to 2, but not equalling 2. It appears that it will always be less than 2.
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